Question
Download Solution PDFIf \(\rm \frac{\sin x-\cos x}{\sin x+\cos x}=\frac{2}{5}\), then the value of \(\rm \frac{1+\cot^2x}{1-\cot^2x}\) is:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFCalculation:
\(\rm \frac{\sin x-\cos x}{\sin x+\cos x}=\frac{2}{5}\)
⇒ \(\rm \frac{\sin x+\cos x}{\sin x-\cos x}=\frac{5}{2}\)
Applying C & D rule, we get
⇒ \(\rm \frac{\sin x}{\cos x}=\frac{5 + 2}{5 -2}\)
⇒ tan x = 7/3
⇒ cot x = 3/7
Now, the value of
\(\rm \frac{1+\cot^2x}{1-\cot^2x}\)
⇒ \(\rm \frac{1+(\frac{3}{7})^2}{1-(\frac{3}{7})^2}\)
⇒ 58/40
⇒ 1.45
∴ The correct answer is 1.45
Last updated on Jun 13, 2025
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